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Find an equation of the straight line tangent, at the given point, to the level curve of the given function passing through that point.

f(x, y) = x^{2}- y^{2} at (2, -1)

1 Answer

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f(2,-1)=2^2-(-1)^2=3

so we're considering the level curve


x^2-y^2=3

The tangent line to this curve at (2, -1) will be the value of
(\mathrm dy)/(\mathrm dx) at this point. Differentiating yields


2x-2y(\mathrm dy)/(\mathrm dx)=0\implies(\mathrm dy)/(\mathrm dx)=\frac xy

and so the slope of the tangent would be
\frac2{-1}=-2.

The tangent line then has equation


y-(-1)=-2(x-2)\implies y=-2x+3
User Jignesh Ansodariya
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